Use LBI/LBO function for extra voltage output

Battery-powered portable equipment are usually designed with step-up-converter ICs that provide one output with a fixed or adjustable voltage. A low-battery-in/low-battery-out (LBI/LBO) function is sometimes incorporated into the chip to monitor a low-battery condition and to warn users when the battery is about to go dead. This Design Idea makes use of this function for extra voltage output.

The Maxim MAX756 boost converter provides a fixed output of 3.3 or 5V at 300 and 200mA, respectively (Figure 1). The input voltage can range from 0.7-5.5V. For low-battery detection, the part has on-chip circuitry comprising a comparator, a reference and an open-drain MOSFET. When the voltage at the LBI input is lower than its threshold level of 1.25V, the MOSFET at the LBO output sinks current to ground.



You can use these components to make a second output with a regulated voltage (Figure 2). R1 and R2 determine the secondary output voltage according to the following equation: Output 2 =VREF(R1+R2)/R2, where VREF is the reference voltage, which is 1.25V for this chip.



You can set Output 2 from 1.25 to 5V as long as it is less than Output 1. Because Output 2 is derived from Output 1, the total output current for both outputs should not exceed 200 and 300mA for Output 1 and Output 2 voltages of 5 and 3.3V, respectively.

You can also use the LBI/LBO function to make a second boost converter (Figure 3). The CD4093 quad Schmitt-triggered NAND gates, inductor L2, R2 through R4, Q1, D1, C1 and C2 compose this boost converter. Add C1 and R2 to IC1B to make a free-running oscillator that IC1A gates on. For the values of R2 and C1 in the figure, the oscillator frequency is approximately 17kHz. R1 pulls up the open-drain LBO output.



When the voltage at the LBI pin is lower than 1.25V, the LBO pin is low, thus allowing operation of the IC1B oscillator. IC1C and IC1D drive power MOSFET transistor Q1. When Q1 is on, it pulls current from inductor L1. When Q1 is off, this energy charges capacitor C2 through flyback diode D1. You apply feedback with resistor divider R3 and R4 to determine the Output 2 voltage, according to the following equation: Output 2=1.25V×(R3+R4)/R4. IC1 gets power from Output 1.



The voltage at Output 2 is a function of the output current and the input voltage (Figure 4). If you have adequate input voltage, the output graph shows a flat section where the IC’s regulation is effective (figures 5 and 6).